Explanation: For a sine wave, the effective voltage is known as the Root Mean Square (RMS) voltage. The relationship between the peak voltage ($V_p$) and the RMS voltage ($V_{RMS}$) is fundamental: $V_{RMS} = V_p / \sqrt{2}$. Knowing that $\sqrt{2}$ is approximately 1.414, we can say that $V_p = V_{RMS} \times 1.414$. Therefore, if the effective voltage is 1 unit, the peak voltage is 1.414 units. This establishes the ratio of peak-to-effective voltage as 1.414 to 1, making option A correct. Conversely, we can express the relationship as $V_{RMS} = V_p \times (1/\sqrt{2})$. Since $1/\sqrt{2}$ is approximately 0.707, this means $V_{RMS} = V_p \times 0.707$. If the peak voltage is 1 unit, the effective voltage is 0.707 units. This establishes the ratio of peak-to-effective voltage as 1 to 0.707, making option B also correct. Since both options A and B accurately describe the same voltage ratio for a sine wave, C is the correct answer.